Question

1.The mean systolic blood pressure for a population of patients (µ) from a local clinic is 130 with a standard deviation (σ) of 18. (You will use the above information for several of the following questions) What is the z-score for a patient with a systolic blood pressure of 123? Rounded to the nearest hundredth.

2.71

-0.83

-0.39

-2.75

-1.14

2.

From information on a previous question: The mean systolic blood pressure for a population of patients (µ) from a local clinic is 130 with a standard deviation (σ) of 18.

What is the probability that a patient chosen at random will have a systolic blood pressure of 123 or below? Rounded to the nearest ten thousandth (4 places to the right of the decimal).

Group of answer choices

0.2033

0.3483

0.2192

0.1587

0.2005

Answer 1

Solution :

Given that ,

1.

mean = = 130

standard deviation = = 18

x = 123

z = x - / = 123 - 130 / 18 = -0.39

z-score = -0.39

2.

mean = = 130

standard deviation = = 18

P(x < 123) = P[(x - ) / < (123 - 130) / 18]

= P(z < -0.39)

= 0.3483

Probability = 0.3483